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%% symmetryBreaking.tex
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%% Made by Alex Nelson
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%% Started on  Sat Nov 29 13:34:02 2008 Alex Nelson
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\section{Spontaneous Symmetry Breaking}

Spontaneous symmetry breaking occurs whenever a given field in a given
Lagrangian has a nonzero vacuum expectation value. Why exactly is this
``breaking'' the symmetry? Well, the Lagrangian appears symmetric
under a symmetry group, but its vacuum state fails to be
symmetric. The system no longer behaves symmetrically. So we went from
symmetry to no symmetry due to a nonzero vacuum expectation value. It
came about from condensed matter physics (see~\cite{wen} for
applications of it in condensed matter physics) but has
since been applied to quantum field theory and particle physics
(see~\cite{peskinSchroeder} for examples in particle physics).

Consider the scalar Lagrangian given by
\begin{equation}
\mathcal{L} =
\underbrace{\frac{1}{2}(\partial_{\mu}\phi)^2}_{\text{``kinetic term''}} +
\underbrace{\frac{1}{2}\mu^{2}\phi^{2} -
  \frac{\lambda}{4!}\phi^{4}}_{\text{``potential term''}}
\end{equation}
where $\phi$ is the scalar field, $\mu$ is a sort of ``mass''
parameter, and $\lambda$ is the coupling. Observe that there is a
symmetry of $\phi\to-\phi$ (a discrete symmetry).  We can think of the
potential as being
\begin{equation}
V(\phi) = -\frac{1}{2}\mu^{2}\phi^2 + \frac{\lambda}{4!}\phi^{4}
\end{equation}
which has extrema when its derivative is zero. There are two, given by
\begin{equation}
\phi_{0} = \pm v = \pm \mu\sqrt{\frac{6}{\lambda}}
\end{equation}
where the constant $v$ is the ``\textbf{vacuum expectation value}''.

We can then write
\begin{equation}
\phi(x) = v + \sigma(x)
\end{equation}
and then rewrite the Lagrangian as
\begin{equation}
\mathcal{L} = \frac{1}{2}(\partial_{\mu}\sigma)^{2} -
\frac{1}{2}(2\mu^2)\sigma^2 - \sqrt{\frac{\lambda}{6}}\mu\sigma^3 -
\frac{\lambda}{4!}\sigma^{4}
\end{equation}
where we dropped the constant terms. We see that the symmetry
$\phi\to-\phi$ is no longer identifiable. 
